up&coming1
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does anyone have any idea of another way to calculate voltage drop besides : 2 x 12.9 (copper) x distance x amperge/ cmils ?
21.2 ( aluminum)
21.2 ( aluminum)
voltage drop calculations
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Carultch
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Make sure to replace "2" with sqrt(3) or 1.732, when dealing with 3-phase.
Bear in mind that those effective "resistivity" values of 12.9 for copper and 21.2 for aluminum are only valid for DC. And likely only at a certain operating temperature, of which I'm not sure. Chapter 9, Table 8 has 75-C resistance values (ohms/kft) for each size, which I typically use. For AC of any size measured in AWG, the distinction doesn't matter, and you can get away with using DC resistance calculations for all practical purposes. For larger sizes (specified by aughts or KCMIL), inductive factors come in to play for AC, and whether the conduit is steel or non-steel may also matter.
Anixter has a nice table with this data, although I find it peculiar that they omit size #3. I did a linear interpolation in terms of 1/kcmil for that size in my lookup table.
https://www.anixter.com/content/dam/Anixter/Guide/7H0011X0_W&C_Tech_Handbook_Sec_07.pdf
Another thing to realize is that the NEC standard tables are based on 75C, which is conservative for most applications. If you've upsized above the minimum local size, obviously the operating temperature should be lower than 75C. So in that event, to be more ambitious (as in opposite of conservative) with this calculation, I think it is reasonable to use Anixter's 60C resistance values.
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Carultch
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This factor represents the effective round trip distance. For single phase circuits, current has to travel the entire distance of twice the 1-way feeder length. For three-phase circuits, the current travels out on one phase, and back, shared upon on the other two, while it is also travelling out on the other two. If you think in terms of phase-to-neutral, the current per phase travels one way on the live conductor, driven to flow by the voltage to neutral. The other two wires carry the same current at a different timing, and undergo the same condition. The neutral doesn't carry current when it is balanced.
If you intend to divide by the phase-to-phase voltage, you use sqrt(3). If you intend to divide by the phase-to-neutral voltage for a 3-phase load on a standard WYE system, you omit this factor altogether (because it is embedded in the calculation of phase-to-neutral voltage). And you'll get the same result.
Example: 25A three phase on a 120/208 system, 50 ft one way on #8 wire.
(25A*50ft*sqrt(3))/((0.778 ohm/kft)*(1000 ft/kft)*208V) = 0.81%
(25A*50ft)/((0.778 ohm/kft)*(1000 ft/kft)*120V) = 0.81%
Julius Right
Senior Member
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- Electrical Engineer Power Station Physical Design Retired
According to:
IEEE Std 141-1993 3.11 Calculation of voltage drops 3.11.1 General mathematical formulas
The approximate formula for the voltage drop is:
V = IRcosf + IX sinf
actual voltage drop= eS+ I*R*cos(f)+ I*X* sin(f)-sqrt( eS^2-( I*X*cos(f)- I*R* sin(f))^2).
where:
V is the voltage drop in circuit, line to neutral
I is the current flowing in conductor
R is the line resistance for one conductor, in ohms
X is the line reactance for one conductor, in ohms
f is the angle whose cosine is the load power factor
cosf is the load power factor, in decimals
sinf is the load reactive factor, in decimals
The voltage drop V obtained from this formula is the voltage drop in one conductor, one way, commonly called the line-to-neutral voltage drop.
For single-phase system multiply by 2.For three-phase system multiply by 1.732
Is it boring,does it?:weeping:
IEEE Std 141-1993 3.11 Calculation of voltage drops 3.11.1 General mathematical formulas
The approximate formula for the voltage drop is:
V = IRcosf + IX sinf
actual voltage drop= eS+ I*R*cos(f)+ I*X* sin(f)-sqrt( eS^2-( I*X*cos(f)- I*R* sin(f))^2).
where:
V is the voltage drop in circuit, line to neutral
I is the current flowing in conductor
R is the line resistance for one conductor, in ohms
X is the line reactance for one conductor, in ohms
f is the angle whose cosine is the load power factor
cosf is the load power factor, in decimals
sinf is the load reactive factor, in decimals
The voltage drop V obtained from this formula is the voltage drop in one conductor, one way, commonly called the line-to-neutral voltage drop.
For single-phase system multiply by 2.For three-phase system multiply by 1.732
Is it boring,does it?:weeping:
2 (or 1.732) x Resistance (ohms/kft) x current (I) x Distance (feet)
R.I.D. method for short. Use NEC Chapter 9, Table 9 (or 8, if applicable) for Ohms to Neutral per 1000 feet for R values.
sparks1
Senior Member
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- Massachusetts
Ed=KIL/CSA
Eg
Table 8
K= 10.4 for #14
#14 =4110 cir mils
Plug in the senario you want
The total length is out and back hot to neutral
Single phase or or three it phase doesn't mater
It's the length , wire size, temp and impedance
will all effect the ED
Table 8 is DC resistance values. Use Table 9 for AC... adjust value for power factor.
Where'd you get the K= 10.4 for #14?
FWIW...
...where CSA units are circular mils.
All voltage drop formulas amount to:
Fitzdrew516
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- Cincinnati, OH
If it's single phase and for some reason you don't want to use the normal voltage drop calc. method then you can use Ohm's law. Any particular reason you don't want to use the 2KID/CMIL method?
As I mentioned earlier, all voltage drop calculations are just a variation on Ohm's Law.
Carultch
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Well, it's all V=I*R in some form or another. It's just a matter of how one determines the value of R, that makes the difference in the methods.
Common misconception. V=I*R is not Ohm's law, but rather a way to write the definition of resistance. R=V/I is the formula that shows this definition directly. For our purposes though, it still works as a shortcut to call this Ohm's law.
Georg Ohm's actual law, is that resistance is independent of voltage and current, which therefore makes the relationship between V and I linear. Meaning, an ordinary material doesn't behave like a diode or a varistor, that can classify as insulators at one voltage and conductors at another voltage. Diodes and Varistors have a variation of resistance as a function of the voltage, and thus are components that operate outside of Ohm's law.
sparks1
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kingpb
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Use the Southwire App
sparks1
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E=I x R is used for DC circuit or for true power calcs not for
AC since we are dealing with XL and it's net effect. Since impedance is present in AC only then it must be considered as part of voltage drop calculation
The question was...
And your reply is...???
Do you realize that in most scientific circles (i.e. non-electrical) that resistivity is given for materials at 20°C?
Approximate K value for #14 uncoated copper in AC circuit, nonmetallic wiring method, at 75°C, using Table 9 value for Effective Z at 0.85 PF.
K = 2.7 ohms/kft × 4.11kcmil = 11.10
I'll leave figuring out K at 60°C as an exercise for you.
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